1.Halley's comet moves about the sun in an elliptical orbit with its closest approach to the sun being 0.59 A.U. and its furthest distance being 35 A.U. [1 Astronomical Unit (A.U.) is the Earth-Sun distance.] If the comet's speed at closest approach is 54 km/s, what is its speed when it is farthest from the Sun?
2.What is the angular momentum of the moon about the Earth? The mass of the moon is 7.35 10 22 kg, the center-to-center separation of the Earth and the Moon is 3.84 10 5 km, and the orbital period of the moon is 27.3 days. Ignore the small offset of the center of mass of the system from the center of the Earth in your calculation.
3. A regulation basketball has a 25.0-cm diameter and a mass of 0.560 kg. It may be approximated as a thin spherical shell with a moment of inertia MR 2. Starting from rest, how long will it take a basketball to roll without slipping 4.00 m down an incline at 30.0 to the horizontal?
4. A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is MR 2, how far will the coin roll up the inclined plane?
I will be happy to check your work, or critique your thinking.
1. To find the comet's speed when it is farthest from the Sun, we can use the concept of conservation of angular momentum.
Angular momentum is given by the equation: L = mvr, where L is the angular momentum, m is the mass, v is the velocity, and r is the distance from the center of rotation (in this case, the Sun).
Since the comet's closest approach is 0.59 A.U., its distance from the Sun can be calculated as r1 = 0.59 * AU.
Also, we know the comet's speed at closest approach is 54 km/s, so v1 = 54 km/s.
To find the comet's speed when it is farthest from the Sun, we can use the conservation of angular momentum equation:
L1 = L2
m * v1 * r1 = m * v2 * r2
Assuming that the mass of the comet remains constant throughout its orbit, we can cancel out the mass term:
v1 * r1 = v2 * r2
Substituting the given values:
(54 km/s) * (0.59 A.U.) = v2 * (35 A.U.)
54 km/s * 0.59 = v2 * 35
v2 = (54 km/s * 0.59) / 35
v2 ≈ 0.9086 km/s
Therefore, the comet's speed when it is farthest from the Sun is approximately 0.9086 km/s.
2. The angular momentum of the moon about the Earth can be calculated using the formula:
L = m * v * r
Where L is the angular momentum, m is the mass of the moon, v is the velocity of the moon, and r is the distance between the center of the Earth and the moon.
Given:
Mass of the Moon (m) = 7.35 × 10^22 kg
Distance between the Earth and the Moon (r) = 3.84 × 10^5 km (convert to meters by multiplying by 1000)
Orbital period of the Moon = 27.3 days (convert to seconds by multiplying by 24 * 60 * 60)
First, we can calculate the velocity of the moon using the formula for circular motion:
v = (2 * π * r) / T
Where T is the orbital period of the moon.
v = (2 * π * 3.84 × 10^8 m) / (27.3 * 24 * 60 * 60 s)
Now we can calculate the angular momentum:
L = (7.35 × 10^22 kg) * (velocity) * (3.84 × 10^8 m)
Calculate the velocity and multiply it by the mass and distance to get the angular momentum.
3. To calculate the time it takes for the basketball to roll without slipping, we can use the concept of rotational motion and the work-energy theorem.
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
The work done on the basketball is equal to the gravitational potential energy it gains while rolling down the inclined plane.
The gravitational potential energy is given by the equation: PE = mgh, where m is the mass of the basketball, g is the acceleration due to gravity, and h is the vertical distance traveled.
Given that the basketball rolls 4.00 m horizontally down an incline at 30.0 degrees to the horizontal, we can calculate the vertical distance traveled as h = 4.00 m * sin(30.0 degrees).
Since the basketball is rolling without slipping, the velocity of the center of mass v_CM is related to the angular velocity ω by the equation: v_CM = ω * r, where r is the radius of the basketball.
The kinetic energy of the basketball can be expressed as: KE = (1/2) * I * ω^2, where I is the moment of inertia of the basketball.
For a thin spherical shell, the moment of inertia can be simplified to I = (2/3) * m * r^2.
Since the basketball starts from rest, its initial kinetic energy is zero.
Therefore, the work done on the basketball is equal to the change in kinetic energy:
mgh = (1/2) * (2/3) * m * r^2 * ω^2
To find the time taken, we can substitute v_CM = ω * r and h = 4.00 m * sin(30.0 degrees) into the equation for work done:
m * g * 4.00 m * sin(30.0 degrees) = (1/2) * (2/3) * m * r^2 * (v_CM / r)^2
Simplify and solve for v_CM:
v_CM = sqrt((3/4) * g * 4.00 m * sin(30.0 degrees) * (3/r))
Finally, we can divide the horizontal distance traveled by the velocity of the center of mass to find the time taken:
Time = 4.00 m / v_CM
4. To calculate the distance the coin rolls up the inclined plane, we can use the concept of conservation of mechanical energy.
The total mechanical energy of the coin is the sum of its kinetic energy (KE) and its gravitational potential energy (PE). Since the coin rolls in a straight line without slipping, the initial kinetic energy is given by KE_initial = (1/2) * I * ω_initial^2, where I is the moment of inertia of the coin and ω_initial is the initial angular speed.
The final potential energy of the coin is given by PE_final = m * g * h, where m is the mass of the coin, g is the acceleration due to gravity, and h is the vertical distance traveled.
The total mechanical energy is conserved, so:
KE_initial + PE_initial = PE_final
Substituting the given values, we get:
(1/2) * I * ω_initial^2 + 0 = m * g * h
Simplifying, we have:
ω_initial^2 = (2 * m * g * h) / I
We can determine the moment of inertia of the coin using the formula for a solid disk:
I = (1/2) * m * r^2, where r is the radius of the coin.
Substituting this value into the equation for ω_initial^2, we have:
ω_initial^2 = (4 * g * h) / r^2
To find the distance the coin rolls up the inclined plane, we can use the equation for displacement:
s = θ * r, where θ is the angle of inclination and r is the radius of the coin.
Since the coin rolls without slipping, the angle of inclination θ can be related to the angular displacement θ_displacement by the equation:
θ_displacement = (s * 180 degrees) / (π * r)
Using this relationship, we can calculate the angular velocity at the final position:
ω_final = ω_initial - (α * t), where α is the angular acceleration and t is the time taken.
The angular acceleration can be calculated using the torque equation:
τ = I * α
Since there is no external torque acting on the coin, the net torque is zero, giving us:
τ_gravity = τ_friction
m * g * r * sin(θ_displacement) = (1/2) * m * r^2 * α
Simplifying, we find:
α = (2 * g * sin(θ_displacement)) / r
Finally, we can substitute the values of ω_initial, ω_final, α, and t into the equation:
θ_displacement = (ω_initial + ω_final) * t / 2
Solving for t, we can calculate the time taken:
t = (2 * θ_displacement) / (ω_initial + ω_final)
Using the calculated time and the equation for displacement, we can find the distance the coin rolls up the inclined plane:
s = θ_displacement * r